## Retaining structures

### Pressure on retaining structures

Liquid pressure

From diving experience we know that the pressure in a liquid (e.g. water) becomes greater the deeper we dive. Consider the vertical surface A-B of the wall in Figure 5. The cube situated at depth h exerts a pressure of w × h (kN) on all its surfaces. If one face of the cube is touching the wall it will exert a pressure onto the wall.

The density of water, w = 1000 kg/m³ = 1Mg/m³
The weight of 1000 kg mass is 1000 × 9.81 = 9810 N (weight = mass × gravitational acceleration). Hence the unit weight of water, w 9.81 kN/m³. Instead of using the exact figure for the gravitational acceleration we approximate this figure to 10 m/s². This provides a safety margin of approximately of 2% and on the other hand it suits the decimal system and eases the calculations. w is the equivalent density of water which in our case, will always be 10 kN/m³. In the right diagram of Figure 1 the pressure at the surface is zero and at depth h is w × h. The average pressure of the 'wetted area' between A and B is ½ w × h (in kN per unit area).

Figure 5

Position of the resulting force

The opposite diagram shows the centre of gravity of a triangle in relation to the principle axis. The centre of gravity for all triangles is at a 1/3rd from the base. (Any side of the triangle can be the base.) A parallel line from the base at 1/3rd of the height of the triangle divides the area into two equal parts (A1=A2). Referring to the above example the resultant force, which equals ½ × w h² (kN) will therefore act at a point of 1/3rd of height from the base.

Example 1

A dam retains water on its vertical surface as shown in Figure 6. The dam has a height of 4.5 metre, and the water level is 0.9 metre below the top of the dam.
What is the resultant water pressure per metre run of dam?

Solution
The equivalent density w of water equals 10 kN/m³

P = ½ × w × h²
= ½ × 10 × 3.62
Figure 6= 64.8 kN

#### Soil pressure (horizontal)

There is some similarity between the calculation of the lateral pressure in water and soil. However, it is apparent that pressures on vertical surfaces from retained soils cannot be determined with the same accuracy as with water. Soils vary in character and weight and behave quite differently under these varying conditions. There are a number of soil pressure theories for the calculation of soil pressure but only the Rankine's theory will be dealt with.
As the density of soil can vary there will be a number of different unit weight figures and not only one as for water. Instead of using w as for the density of water, will be used for the density of soil (see also properties of soil).

Consider a mass of soil with a horizontal upper surface. If the unit weight of the soil is then an element at a depth h below the surface will be subjected to a vertical pressure g × h. This stress is a major principal stress is, i.e.1 = g h (multiplying sign omitted). There is of course also a lateral stress or minor principal stress 3. The ratio between the 1/3 for Figure 7 a soil at rest is given the symbol
Ko
and is called the coefficient of earth pressure at rest. The lateral pressure in a soil at rest is equal to Ko × h

#### Angle of repose

Consider, for example, the soil retained by the vertical face AB in Figure 8. If the wall (retaining face AB) was removed, then some of the soil would probably collapse. After the soil has collapsed it would assume a line BC as shown. The angle made between the horizontal and the line BC will vary with different types of soil. This angle is called the Angle of repose or the angle of internal friction of the soil.
Figure 8

#### Active and passive earth pressure

Contemplate a smooth vertical wall supporting a mass of soil at rest in which the lateral pressure on the wall is = Ko h. If the wall is allowed to yield, i.e. move forward slightly, there will be an immediate reduction in the value of lateral stress, but if the wall is pushed slightly into the soil there will be an increase in the value of the lateral pressure.
The minimum value is known as the active earth pressure (Ea), and the value equals Ka × h where Ka) = the coefficient of active earth pressure.

Rankine's theory states in general terms that the active earth pressure coefficient is:

The active earth pressure at a depth h (m) due to a level fill of soil is therefore:

The passive earth pressure (Ep), which equals Kp× h where Kp = the coefficient of passive earth pressure.

#### Example 2

A soil weighing 19 kN/m3 and having an angle of repose of 34°, exerts a pressure on a 3.6 metre high vertical face of a wall.

What is the resultant horizontal force per metre run of wall?

Figure 9

Ea = 0.283 × 19 × 3.6 = 19.34 kN/m

The total horizontal force from the retained earth acting on the wall surface is:

P = 19.34 × ½ 3.6 = 34.84 kN

It is of advantages to calculate the individual components, especially if you need to know the Ea figure. Compare this result with the liquid pressure in Example 1. As you can see the force due to hydrostatic (liquid) pressure is much less than the force due to earth pressure. This results from the internal friction ( higher in sand than in clay) between the grains of the soil represented by the Ka figure.